Step 1: Understanding the Concept:
This problem involves analyzing the motion of a freely falling body. The distance covered by a freely falling body is not linear with time; it is proportional to the square of the time. We need to find the distance fallen in time T/2 and relate it to the total height H.
Step 2: Key Formula or Approach:
The equation for the distance covered by a body starting from rest under constant acceleration is:
\[ s = ut + \frac{1}{2}at^2 \]
For a body released from rest, $u=0$, and the acceleration is $a=g$. So, the distance fallen from the top is $s = \frac{1}{2}gt^2$.
Step 3: Detailed Explanation:
Let's analyze the motion for the total time T.
The body falls a total height H in time T. Using the kinematic equation:
\[ H = (0)T + \frac{1}{2}gT^2 \implies H = \frac{1}{2}gT^2 \quad \dots(1) \]
Now, let's find the distance the body has fallen from the top at time $t = T/2$. Let's call this distance $h$.
\[ h = \frac{1}{2}g t^2 = \frac{1}{2}g \left(\frac{T}{2}\right)^2 = \frac{1}{2}g \frac{T^2}{4} = \frac{1}{8}gT^2 \]
We want to express $h$ in terms of H. From equation (1), we know that $\frac{1}{2}gT^2 = H$.
Substitute this into the expression for $h$:
\[ h = \frac{1}{4} \left(\frac{1}{2}gT^2\right) = \frac{H}{4} \]
This is the distance fallen from the top. The question asks for the position of the body from the ground.
Position from ground = Total height - Distance fallen from top
\[ \text{Position from ground} = H - h = H - \frac{H}{4} = \frac{3H}{4} \]
Step 4: Final Answer:
At time T/2, the body is at a height of $\frac{3H}{4}$ from the ground. Therefore, option (C) is correct.